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SOLUTION OF HW8 MINGFENG ZHAO November 06, 2011 1. Show that lim f (x) = A if and only if x→p lim [f (x) − A] = 0. x→p Proof. (=⇒) If lim f (x) = A, then for any > 0, there exists some δ > 0 such that x→p |f (x) − A| < , for all |x − p| < δ. That is, lim [f (x) − A] = 0. x→p (⇐=) If lim [f (x) − A] = 0, then for any > 0, there exists some δ > 0 such that x→p |f (x) − A| < , for all |x − p| < δ. That is, lim f (x) = A. x→p 2. If lim f (x) = A, and lim g(x) = B, show that x→p x→p lim [f (x) − g(x)] = A − B. x→p Proof. Since lim f (x) = A, then for any > 0, there exists some δ1 > 0 such that x→p |f (x) − A| < , 2 for all |x − p| < δ1 . For the above > 0, since lim g(x) = B, then there exists some δ2 > 0 such that x→p |g(x) − B| < , 2 for all |x − p| < δ2 . 1 2 MINGFENG ZHAO Now taking δ = min {δ1 , δ2 } > 0, for all |x − p| < δ, we have |f (x) − A| < , 2 and |g(x) − B| < . 2 Then we have |[f (x) − g(x)] − [A − B]| = |f (x) − A + g(x) − B| ≤ |f (x) − A| + |g(x) − B| < + 2 2 = , for all |x − p| < δ. That is, lim [f (x) − g(x)] = A − B. x→p 3. Show that lim x3 = p3 for all p ∈ R x→p Proof. For any p ∈ R and fix, for any |x−p| < 1, we know that |x| = |x−p+p| ≤ |x−p|+|p| < 1+|p|. For any > 0, we know that for all |x − p| < 1, we have |x3 − p3 | = Let δ = min |(x − p)(x2 + xp + p2 )| = |x − p||x2 + xp + p2 | ≤ |x − p|[|x|2 + |x||p| + |p|2 ] ≤ |x − p|[[1 + |p|]2 + (1 + |p|)|p| + |p|2 ]. 1, [[1 + |p|]2 + (1 + |p|)|p| + |p|2 ] |x3 − p3 | > 0, then for all |x − p| < δ, we have ≤ |x − p|[[1 + |p|]2 + (1 + |p|)|p| + |p|2 ] < δ · [[1 + |p|]2 + (1 + |p|)|p| + |p|2 ] ≤ δ· [[1 + |p|]2 + (1 + |p|)|p| + |p|2 ] SOLUTION OF HW8 = 3 . So we get |x3 − p3 | < , for all |x − p| < δ. That is, lim x3 = p3 x→p 4. [Problem 22, in Page 138] Let f (x) = sin x, if x ≤ c ax + b, if x > c. Where a, b, c are constants. If b and c are given, and find all values of a for which f is continuous at the point x = c. Proof. By the definition of f , we know that f (c+) = lim f (x) = lim [ax + b] = ac + b. x&c x&c And f (c−) = lim f (x) = lim sin x = sin c. x%c x&c Since f is continuous at x = c, then f (c+) = f (c−), that is, we have ac + b = sin c. If c = 0, then we must have b = 0, that is, if b = c = 0, for all a ∈ R, then f is continuous at x = c. If c 6= 0, then we have a = sin c − b . c Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: [email protected]